calculate-1-1-x-4-x-2-1-2-e-x-e-x-1-dx-

Question Number 57421 by Abdo msup. last updated on 03/Apr/19

c a l c u l a t e \int_{- 1}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2} + e^{x}}{e^{x} + 1} d x

Answered by einsteindrmaths@hotmail.fr last updated on 04/Apr/19

= \int_{- 1} \overset{1}{} \frac{{(x^{4} + x^{2} + 1)}^{2} + e^{x}}{e^{x} + 1} d x = \int_{- 1} \overset{1}{} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x + \underset{- 1}{\int^{1}} \frac{e^{x}}{e^{x} + 1} d x

t h e s e c o n d e i n t e g r a l i s e a s y t o e v a l u a t

\int \frac{e^{x}}{e^{x} + 1} d x = l n (e^{x} + 1) + c

l e t s f i n d \underset{- 1}{\int^{1}} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x = \underset{- 1}{\int^{0}} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x + \underset{0}{\int^{1}} \frac{{(x^{4} + x^{2} + 1)}^{4}}{e^{x} + 1} d x

w e p u t y = - x i n t h e f i r s t e w e g e t \int_{0}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{- y} + 1} d y + \int_{0}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x

= \underset{0}{\int^{1}} [\frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} + \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{- x} + 1}] d x = \int_{0}^{1} {(x^{4} + x^{2} + 1)}^{2} d x = \underset{0}{\int^{1}} (x^{8} + 2 x^{6} + 3 x^{4} + 2 x^{2} + 1) d x

e a s y t o e v a l u a t